Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

What is the net force (magnitude and direction) on the electron moving in the magnetic field in Fig. 1.10 if B = 2 T, v = 4 x 10⁴ m/s, and $\theta$ = 30 ^o ?

  

Figure 1.10: Point charge in a magnetic field

Solution:
We first find the magnitude of the force as

F = qvBsin $$\theta$$ = 1.6 x 10^{- 19} $$\cdot$$ 4 x 10⁴ $$\cdot$$ 2 $$\cdot$$ sin 30 ^o = 6.4 x 10^{- 15} N .

To determine the direction, we note that $\vec{v}$ x $\vec{B}$ is directed out of the page, and so the force on a negative charge is opposite to this. Thus, the force on the electron is 6.4 x 10^{- 15} N directed into the page.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose a very long straight wire of linear mass density 20 g/m is immersed in a constant magnetic field B = 3 T, as in Fig. 1.11. What current I would be required so that the wire will be suspended?

  

Figure 1.11: Long straight wire in a magnetic field

Solution:
We note that the force per unit length due to gravity,

$${\frac{F_g}{l}}$$ = $${\frac{m}{l}}$$g,

is directed downwards, while the force per unit length due to the magnetic field,

$${\frac{F_B}{l}}$$ = IB,

is acting upwards. In order that they balance, we demand

$${\frac{m}{l}}$$g = IB $$\Rightarrow$$ I = $${\frac{m}{l}}$$$${\frac{g}{B}}$$ = $${\frac{0.02\times 9.8}{3}}$$ = 0.065 A .

Thus, the wire must carry a current of 0.065 A in order to be suspended.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Consider the mass spectrometer in Fig. 1.12. The electric field between the plates of the velocity selector is E = 950 V/m, and the magnetic field B in both the velocity selector and in the deflection chamber has a magnitude of 0.9 T. Find the radius r for a singly charged ion of mass m = 2.18 x 10^{- 26} kg in the deflection chamber.

  

Figure 1.12: Mass spectrometer

Solution:
We first use the fact that in order for the ion to pass through the velocity selector undeflected the force due to the electric and magnetic fields must balance. Note that the force on a positive charge is downwards due to the electric field and upwards due to the magnetic field; to balance, we then have

qE = qvB $$\Rightarrow$$ v = $${\frac{E}{B}}$$.

When the ion enters the deflection chamber it experiences a magnetic force, causing it to go in a circular orbit. The magnetic force then gives rise to the centripetal acceleration as

F = qvB = m$${\frac{v^2}{r}}$$ $$\Rightarrow$$ qB = $${\frac{mv}{r}}$$.

We thus find

r = $${\frac{m}{qB}}$$v = $${\frac{mE}{qB^2}}$$ = $${\frac{2.18\times 10^{-26}\cdot 950}{1.6\times 10^{-19}\cdot (0.9)^2}}$$ = 1.6 x 10^{- 4} m .

Thus, the radius of the orbit is 0.16 mm.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The two long straight wires in Fig. 1.13 each carry a current of I = 5 A in opposite directions and are separated by a distance d = 30 cm. Find the magnetic field a distance l = 20 cm to the right of the wire on the right.

  

Figure 1.13: Magnetic field of two long straight wires

Solution:
We shall use the expression for the magnetic field of a long straight wire a distance r away as

B = $${\frac{\mu_0I}{2\pi r}}$$,

with direction given by the appropriate right-hand-rule. Due to the wire on the right, the magnetic field is

B_R = $${\frac{\mu_0I}{2\pi l}}$$ = $${\frac{4\pi\times 10^{-7}\cdot 5}{2\pi\cdot 0.1}}$$ = 1.0 x 10^{- 5} T ,

which is directed out of the page. Due to the wire on the left, the magnetic field is

B_L = $${\frac{\mu_0I}{2\pi (l+d)}}$$ = $${\frac{4\pi\times 10^{-7}\cdot 5}{2\pi\cdot 0.5}}$$ = 2.0 x 10^{- 6} T ,

which is directed into the page. Thus, the net magnetic field is 1.0 x 10^{- 5} - 2.0 x 10^{- 6} = 8.0 x 10^{- 6} T directed out of the page.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

For the arrangement shown in Fig. 1.14, the long straight wire carries a current of I₁ = 5 A. This wire is a distance d = 0.1 m away from a rectangular loop of dimensions a = 0.3 m and b = 0.4 m which carries a current I₂ = 10 A. Find the net force exerted on the rectangular loop by the long straight wire.

  

Figure 1.14: Force on a current loop due to a long straight wire

Solution:
The force $\vec{F}$ = l$\vec{I}$ x $\vec{B}$ on the rectangular loop arises due to the magnetic field B = $\mu_{0}^{}$I/2$\pi$r of the long straight wire. We first of all note that the force cancels between the top and bottom portions of the loop. Now, on the left portion of the loop,

B_L = $${\frac{\mu_0I_1}{2\pi d}}$$,

directed into the page, which gives rise to a force

F_L = bI₂B_L = $${\frac{bI_2\mu_0I_1}{2\pi d}}$$ = $${\frac{0.4\cdot10\cdot4\pi\times10^{-7}\cdot 5}{2\pi\cdot 0.1}}$$ = 4 x 10^{- 5} N ,

which is directed to the left. On the right portion of the loop,

B_R = $${\frac{\mu_0I_1}{2\pi (a+d)}}$$,

directed into the page, which gives rise to a force

F_R = bI₂B_R = $${\frac{bI_2\mu_0I_1}{2\pi (a+d)}}$$ = $${\frac{0.4\cdot10\cdot4\pi\times10^{-7}\cdot 5}{2\pi\cdot (0.3+0.1)}}$$ = 1 x 10^{- 5} N ,

which is directed to the right. Thus, the net force is 4 x 10^{- 5} - 1 x 10^{- 5} = 3 x 10^{- 5} N directed to the left.